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If the amplitude of $z-2-3 i$ is $\frac{\pi}{4}$, then the locus of $z=x+i y$ is :
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Verified Answer
The correct answer is:
$x-y+1=0$
We have, $z-2-3 i=x+i y-2-3 i$
$=(x-2)+i(y-3)$
Given, $\tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4} \Rightarrow y-3=x-2$
$\therefore \quad x-y+1=0$
$=(x-2)+i(y-3)$
Given, $\tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4} \Rightarrow y-3=x-2$
$\therefore \quad x-y+1=0$
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